The mole concept Questions in English

(N/A) Avogadro constant: The number of particles (atoms,molecules,or ions) present in $1$ mole of a substance is known as the Avogadro constant,which is $6.022 \times 10^{23} \ mol^{-1}$.
Molar volume: According to Avogadro's law,the volume of a gas $(V)$ is directly proportional to the number of moles $(n)$ at constant temperature and pressure. At $STP$,$1$ mole of an ideal gas occupies a volume of $22.71098 \ L \ mol^{-1}$.
$STP$ (Standard Temperature and Pressure): According to current $IUPAC$ recommendations,$STP$ is defined as a temperature of $273.15 \ K$ $(0^{\circ} C)$ and a pressure of $1 \ bar$ $(10^{5} \ Pa)$.

(N/A) At $STP$,$22.4 \ L$ of any gas contains $6.022 \times 10^{23}$ molecules (Avogadro's number).
Number of moles of $CH_4 = \frac{5.6 \ L}{22.4 \ L/mol} = 0.25 \ mol$.
Number of molecules of $CH_4 = 0.25 \times 6.022 \times 10^{23} = 1.5055 \times 10^{23}$ molecules.
Number of $C$ atoms $= 1 \times 1.5055 \times 10^{23} = 1.5055 \times 10^{23}$ atoms.
Number of $H$ atoms $= 4 \times 1.5055 \times 10^{23} = 6.022 \times 10^{23}$ atoms.
Total number of atoms $= 1.5055 \times 10^{23} + 6.022 \times 10^{23} = 7.5275 \times 10^{23}$ atoms.

(A) The number of molecules given is $6.022 \times 10^{23}$,which is equal to $1 \ \text{mole}$ of $O_2$ molecules.
Since $1 \ \text{mole}$ of any gas at $STP$ occupies $22.4 \ L$,the volume is $22.4 \ L$.
The molar mass of $O_2$ is $2 \times 16 = 32 \ g/mol$.
Therefore,the mass of $1 \ \text{mole}$ of $O_2$ is $32 \ g$.

(C) mole is defined as the amount of a substance that contains exactly $6.022 \times 10^{23}$ elementary entities of the given substance.
This number is known as the Avogadro constant $(N_A)$.
It is the $SI$ unit for the amount of substance.

(A) The chemical formula is $Na_2CO_3 \cdot 10H_2O$.
In $1 \ mol$ of $Na_2CO_3 \cdot 10H_2O$,there are $3$ oxygen atoms in $Na_2CO_3$ and $10$ oxygen atoms in $10H_2O$,totaling $13$ moles of oxygen atoms.
For $0.1 \ mol$ of the compound,the moles of oxygen atoms = $0.1 \times 13 = 1.3 \ mol$.
The molar mass of oxygen is $16 \ g/mol$.
Mass of oxygen = $1.3 \ mol \times 16 \ g/mol = 20.8 \ g$.

(A) The molar mass of methane $(CH_4)$ is calculated as: $12 + (4 \times 1) = 16 \ g/mol$.
According to the mole concept,$1 \ mol$ of $CH_4$ contains $6.022 \times 10^{23}$ molecules,which weighs $16 \ g$.
Therefore,the weight of one molecule of $CH_4$ is: $\frac{16}{6.022 \times 10^{23}} \ g$.
Calculation: $16 \div 6.022 \approx 2.657$.
Thus,the weight is $2.657 \times 10^{-23} \ g$.

(A) The Avogadro's number,denoted by $N_A$,represents the number of particles (atoms,molecules,or ions) in one mole of a substance. Its value is $6.022 \times 10^{23} \ mol^{-1}$.

(A) The number of moles $(n)$ is calculated using the formula: $n = \frac{\text{Given Volume}}{\text{Molar Volume at STP}}$.
Given volume $= 0.224 \ L$.
Molar volume at $STP = 22.4 \ L \ mol^{-1}$.
$n = \frac{0.224 \ L}{22.4 \ L \ mol^{-1}} = 0.01 \ mol$.

(A) The mass of $H_2$ in $100 \ g$ of the mixture is $20 \ g$.
The molar mass of $H_2$ is $2 \ g/mol$.
The number of moles of $H_2$ is calculated as: $n = \frac{\text{mass}}{\text{molar mass}} = \frac{20 \ g}{2 \ g/mol} = 10 \ mol$.

(A) The molar mass of water $(H_2O)$ is $18 \ g \ mol^{-1}$.
Assuming the density of water is $1 \ g \ mL^{-1}$,then $1 \ L$ of water has a mass of $1000 \ g$.
The number of moles of $H_2O$ in $1 \ L$ is calculated as:
$n = \frac{1000 \ g}{18 \ g \ mol^{-1}} = 55.55 \ mol$.
Since molarity $(M)$ is defined as moles per liter of solution:
$M = \frac{55.55 \ mol}{1 \ L} = 55.55 \ mol \ L^{-1}$.

(A) The number of atoms is calculated using the formula: $\text{Number of atoms} = \frac{\text{mass}}{\text{molar mass}} \times N_A \times \text{atomicity}$.
$(A)$ For $1 \ g$ of $Li$: $\frac{1}{7} \times N_A \times 1 \approx 0.143 \times N_A$.
$(B)$ For $1 \ g$ of $Ag$: $\frac{1}{108} \times N_A \times 1 \approx 0.009 \times N_A$.
$(C)$ For $1 \ g$ of $Mg$: $\frac{1}{24} \times N_A \times 1 \approx 0.042 \times N_A$.
$(D)$ For $1 \ g$ of $O_2$: $\frac{1}{32} \times N_A \times 2 = \frac{1}{16} \times N_A \approx 0.0625 \times N_A$.
Comparing the values,$1 \ g$ of $Li$ has the maximum number of atoms.

(A) The number of atoms in one mole of a substance is defined by Avogadro's number $(N_A)$.
Given that the mass of one mole of carbon is $12 \; g$ and the mass of a single carbon-$12$ atom is $1.9926 \times 10^{-23} \; g$.
The number of atoms $= \frac{\text{Total mass}}{\text{Mass of one atom}} = \frac{12 \; g}{1.9926 \times 10^{-23} \; g} \approx 6.022 \times 10^{23}$ atoms.
Thus,the number of atoms in one mole of carbon is $6.022 \times 10^{23}$.

(C) The number of moles of sodium is calculated as: $n = \frac{\text{mass}}{\text{molar mass}} = \frac{8 \ g}{23 \ g/mol} \approx 0.3478 \ mol$.
The number of atoms is given by: $N = n \times N_{A} = 0.3478 \times 6.02 \times 10^{23} \approx 2.09 \times 10^{23}$.
Given the expression $x \times 10^{23}$,we have $x \approx 2.09$.
Rounding to the nearest integer,$x = 2$.

(B) The molar mass of $C_{7}H_{5}N_{3}O_{6}$ is $(7 \times 12) + (5 \times 1) + (3 \times 14) + (6 \times 16) = 84 + 5 + 42 + 96 = 227 \ g/mol$.
The number of moles of $C_{7}H_{5}N_{3}O_{6}$ is $n = \frac{681 \ g}{227 \ g/mol} = 3 \ mol$.
Since each molecule of $C_{7}H_{5}N_{3}O_{6}$ contains $3$ atoms of $N$,the total moles of $N$ atoms is $n_{N} = 3 \times 3 = 9 \ mol$.
The number of $N$ atoms is $9 \times N_{A} = 9 \times 6.02 \times 10^{23} = 54.18 \times 10^{23} = 5418 \times 10^{21}$.
Therefore,the value of $x$ is $5418$.

(B) The volume of the spherical droplet is given by $V = \frac{4}{3} \pi r^3$.
Given $r = 1.0 \ cm$,so $V = \frac{4}{3} \pi (1.0)^3 = \frac{4 \pi}{3} \ cm^3$.
The mass of the water droplet is $m = \text{density} \times \text{volume} = 1.0 \ g \ cm^{-3} \times \frac{4 \pi}{3} \ cm^3 = \frac{4 \pi}{3} \ g$.
The molar mass of water $(H_2O)$ is $M = 18 \ g \ mol^{-1}$.
The number of moles $n$ is calculated as $n = \frac{m}{M} = \frac{4 \pi / 3}{18} = \frac{4 \pi}{54} = \frac{2 \pi}{27} \ mol$.

(A) Given: density of water $= 1.0\, g\, mL^{-1}$.
Volume of water $= 250\, mL$.
Mass of water $= \text{density} \times \text{volume} = 1.0\, g\, mL^{-1} \times 250\, mL = 250\, g$.
Molar mass of water $(H_2O) = (2 \times 1) + 16 = 18\, g\, mol^{-1}$.
Number of moles of water $= \frac{\text{Mass}}{\text{Molar mass}} = \frac{250}{18} \approx 13.889\, mol$.
Number of molecules $= \text{moles} \times \text{Avogadro's number} = 13.889 \times 6.023 \times 10^{23} \approx 83.65 \times 10^{23}$.
Thus,the number of water molecules is closest to $83.6 \times 10^{23}$.

(A) In the molecular formula $C_2F_4$,there are $4$ fluorine atoms. The atomic mass of $F$ is $19$,so $4$ fluorine atoms contribute $76$ to the molar mass.
Possible molar masses of $C_2F_4$ depend on the carbon isotopes:
$100$ $(76 + 12 + 12)$,$101$ $(76 + 12 + 13)$,and $102$ $(76 + 13 + 13)$.
Given the abundances: $^{12}C = 99\%$ $(0.99)$ and $^{13}C = 1\%$ $(0.01)$.
The probability of having a molar mass of $101$ corresponds to the combination of one $^{12}C$ and one $^{13}C$ atom.
Since there are two carbon positions,the combinations are $(^{12}C, ^{13}C)$ or $(^{13}C, ^{12}C)$.
Percentage of $C_2F_4$ with molar mass $101 = 2 \times (0.99 \times 0.01) \times 100 = 1.98\%$.

(C) For the $1^{st}$ jar:
Number of moles of $H_2 = \frac{2 \ g}{2 \ g/mol} = 1 \ mol$.
Number of molecules of $H_2 = 1 \times N_A = 6.022 \times 10^{23}$ molecules.
For the $2^{nd}$ jar:
Number of moles of $N_2 = \frac{28 \ g}{28 \ g/mol} = 1 \ mol$.
Number of molecules of $N_2 = 1 \times N_A = 6.022 \times 10^{23}$ molecules.
Since both jars contain $1 \ mol$ of gas,they contain the same number of molecules.

(D) At $STP$,the molar volume of an ideal gas is $22.7 \ L \ mol^{-1}$.
Number of moles of $O_2 = \frac{\text{Volume at } STP}{22.7 \ L \ mol^{-1}} = \frac{2.8375 \ L}{22.7 \ L \ mol^{-1}} = 0.125 \ mol$.
Number of molecules $= \text{Number of moles} \times N_A = 0.125 \times 6.022 \times 10^{23} \ mol^{-1} = 7.5275 \times 10^{22}$ molecules.
Therefore,the values are $7.527 \times 10^{22}$ molecules and $0.125 \ mol$.

(D) $(1)$ $4 \ u$ of $He = \frac{4 \ u}{4 \ u} = 1 \ He$ atom
$(2)$ $4 \ g$ of helium $= \frac{4 \ g}{4 \ g \ mol^{-1}} = 1 \ mole = N_A \ He$ atoms
$(3)$ $2.271098 \ L$ of $He$ at $STP = \frac{2.271098}{22.71098} \ mol = 0.1 \ mol = 0.1 \ N_A \ He$ atoms
$(4)$ $4 \ mol$ of $He = 4 \ N_A \ He$ atoms
Comparing the values,$4 \ mol$ contains the highest number of atoms.

(A) The molar mass of $CO_2$ is $44 \ g/mol$.
Initial moles of $CO_2$ = $\frac{x \times 10^{-3} \ g}{44 \ g/mol} = \frac{x \times 10^{-3}}{44} \ mol$.
Moles of $CO_2$ removed = $\frac{10^{21}}{6.02 \times 10^{23}} \approx 1.66 \times 10^{-3} \ mol$.
Moles left = $\text{Initial moles} - \text{Removed moles}$.
$2.8 \times 10^{-3} = \frac{x \times 10^{-3}}{44} - 1.66 \times 10^{-3}$.
$2.8 \times 10^{-3} + 1.66 \times 10^{-3} = \frac{x \times 10^{-3}}{44}$.
$4.46 \times 10^{-3} = \frac{x \times 10^{-3}}{44}$.
$x = 4.46 \times 44 = 196.24 \ mg$.

(A) $1$. The starting material is $2$-(hydroxymethyl)cyclopentanol. Oxidation with excess $CrO_3$ gives $2$-oxocyclopentanecarboxylic acid $(P)$.
$2$. Protection of the ketone with ethylene glycol gives $(Q)$.
$3$. Reaction with $CH_3MgBr$ followed by acid workup removes the protecting group and gives $2$-oxocyclopentanecarboxylic acid $(R)$ (as the Grignard reagent reacts with the acid group,but the overall sequence regenerates the keto-acid).
$4$. Reduction of $(R)$ with $NaBH_4$ reduces the ketone to an alcohol,yielding $2$-hydroxycyclopentanecarboxylic acid $(S)$.
$5$. The molecular formula of $(S)$ is $C_6H_{10}O_3$.
$6$. Molar mass of $(S) = (6 \times 12) + (10 \times 1) + (3 \times 16) = 72 + 10 + 48 = 130 \ g \ mol^{-1}$.
$7$. Weight of $0.1$ mole of $(S) = 0.1 \ mol \times 130 \ g \ mol^{-1} = 13 \ g$.

(B) The number of atoms is calculated as: $\text{Number of atoms} = \text{Atomicity} \times \text{moles} \times N_A$.
$(A)$ $Na_2CO_3$: $\text{Atomicity} = 6$,$\text{moles} = \frac{212}{106} = 2$. $\text{Atoms} = 6 \times 2 \times N_A = 12 N_A$.
$(B)$ $Na_2O$: $\text{Atomicity} = 3$,$\text{moles} = \frac{248}{62} = 4$. $\text{Atoms} = 3 \times 4 \times N_A = 12 N_A$.
$(C)$ $NaOH$: $\text{Atomicity} = 3$,$\text{moles} = \frac{240}{40} = 6$. $\text{Atoms} = 3 \times 6 \times N_A = 18 N_A$.
$(D)$ $H_2$: $\text{Atomicity} = 2$,$\text{moles} = \frac{12}{2} = 6$. $\text{Atoms} = 2 \times 6 \times N_A = 12 N_A$.
$(E)$ $CO_2$: $\text{Atomicity} = 3$,$\text{moles} = \frac{220}{44} = 5$. $\text{Atoms} = 3 \times 5 \times N_A = 15 N_A$.
Thus,$A, B$ and $D$ have an equal number of atoms $(12 N_A)$.

(B) $1$. Calculate moles of $CO_2$: Molar mass of $CO_2 = 12 + 2 \times 16 = 44 \ g/mol$. Moles of $CO_2 = \frac{4.4 \ g}{44 \ g/mol} = 0.1 \ mol$.
$2$. Calculate moles of $H_2$: At $\text{STP}$,$1 \ mol$ of gas occupies $22.4 \ L$. Moles of $H_2 = \frac{2.24 \ L}{22.4 \ L/mol} = 0.1 \ mol$.
$3$. Total moles of molecules = $0.1 \ mol (CO_2) + 0.1 \ mol (H_2) = 0.2 \ mol$.
$4$. Total number of molecules = $\text{Total moles} \times N_A = 0.2 \times 6.022 \times 10^{23} = 1.2044 \times 10^{23}$.

(B) The number of molecules is directly proportional to the number of moles $(n = \frac{\text{mass}}{\text{molar mass}})$.
$7 \ g$ $N_2 = \frac{7}{28} = 0.25 \ mol$
$2 \ g$ $H_2 = \frac{2}{2} = 1 \ mol$
$16 \ g$ $NO_2 = \frac{16}{46} \approx 0.348 \ mol$
$16 \ g$ $O_2 = \frac{16}{32} = 0.5 \ mol$
Since $2 \ g$ $H_2$ has the highest number of moles $(1 \ mol)$,it contains the maximum number of molecules.

(A) To find the number of atoms,we use the formula: $\text{Number of atoms} = \frac{\text{mass}}{\text{molar mass}} \times N_A \times \text{atomicity}$.
$A$) For $C_4H_{10}$ (Molar mass = $58 \ g/mol$): $\text{Atoms} = \frac{1}{58} \times N_A \times 14 \approx 0.241 \ N_A$.
$B$) For $N_2$ (Molar mass = $28 \ g/mol$): $\text{Atoms} = \frac{1}{28} \times N_A \times 2 \approx 0.071 \ N_A$.
$C$) For $Ag$ (Molar mass = $108 \ g/mol$): $\text{Atoms} = \frac{1}{108} \times N_A \times 1 \approx 0.009 \ N_A$.
$D$) For $H_2O$ (Molar mass = $18 \ g/mol$): $\text{Atoms} = \frac{1}{18} \times N_A \times 3 \approx 0.166 \ N_A$.
Comparing the values,$1 \ g$ of $C_4H_{10}$ contains the maximum number of atoms.

(C) The number of molecules is calculated as: $\text{Number of molecules} = \text{moles} \times N_A = \frac{\text{mass}}{\text{molar mass}} \times N_A$.
For $A$: $44 \ g$ $CO_2$ (molar mass $= 44 \ g/mol$): $\text{moles} = 44/44 = 1 \ mol$.
For $B$: $48 \ g$ $O_2$ (molar mass $= 32 \ g/mol$): $\text{moles} = 48/32 = 1.5 \ mol$.
For $C$: $8 \ g$ $H_2$ (molar mass $= 2 \ g/mol$): $\text{moles} = 8/2 = 4 \ mol$.
For $D$: $64 \ g$ $SO_2$ (molar mass $= 64 \ g/mol$): $\text{moles} = 64/64 = 1 \ mol$.
Since $C$ has the highest number of moles $(4 \ mol)$,it contains the maximum number of molecules.

(C) The number of protons in one $NH_{4}^{+}$ ion is calculated as: $7 (N) + 4 (H) = 11$ protons.
The molar mass of $NH_{4}^{+}$ is $14 + 4 \times 1 = 18 \ g/mol$.
The number of moles of $NH_{4}^{+}$ in $1.8 \ g$ is $\frac{1.8 \ g}{18 \ g/mol} = 0.1 \ mol$.
The number of $NH_{4}^{+}$ ions is $0.1 \times N_{A}$.
Therefore,the total number of protons is $11 \times 0.1 \times N_{A} = 1.1 N_{A}$.

(C) The molar mass of sucrose $(C_{12}H_{22}O_{11})$ is $(12 \times 12) + (22 \times 1) + (11 \times 16) = 144 + 22 + 176 = 342 \ g/mol$.
Number of moles of sucrose = $\frac{\text{Given mass}}{\text{Molar mass}} = \frac{34.2 \ g}{342 \ g/mol} = 0.1 \ mol$.
Since $1 \ mol$ of $C_{12}H_{22}O_{11}$ contains $12 \ mol$ of carbon atoms,
Number of moles of carbon atoms = $12 \times 0.1 \ mol = 1.2 \ mol$.

(C) Step $1$: Calculate the total number of molecules in $44 \ mg$ of $CO_2$.
Molar mass of $CO_2 = 12 + 2 \times 16 = 44 \ g/mol$.
Mass of $CO_2 = 44 \ mg = 44 \times 10^{-3} \ g$.
Number of moles $= \frac{44 \times 10^{-3} \ g}{44 \ g/mol} = 10^{-3} \ mol$.
Total molecules $= 10^{-3} \times 6 \times 10^{23} = 6 \times 10^{20}$ molecules.
Step $2$: Subtract the removed molecules from the total.
Molecules left $= (6 \times 10^{20}) - (2 \times 10^{20}) = 4 \times 10^{20}$ molecules.

(D) The definition of molar mass is the mass of one mole of a substance in grams,not milligrams.
Therefore,the statement in option $D$ is incorrect.

(D) Density of water $(d)$ $= 1 \ g/mL$.
Mass of $0.0018 \ mL$ of water $= 0.0018 \ mL \times 1 \ g/mL = 0.0018 \ g$.
Molar mass of $H_2O = 18 \ g/mol$.
Number of moles $= \frac{0.0018 \ g}{18 \ g/mol} = 10^{-4} \ mol$.
Number of molecules $= \text{moles} \times N_A = 10^{-4} \times 6.023 \times 10^{23} = 6.023 \times 10^{19}$ molecules.

(C) For $16 \ g$ of oxygen $(O_2)$: $\text{Number of atoms} = \frac{16 \ g}{32 \ g/mol} \times 2 \times N_A = N_A$.
For $16 \ g$ of ozone $(O_3)$: $\text{Number of atoms} = \frac{16 \ g}{48 \ g/mol} \times 3 \times N_A = N_A$.
Since both contain $N_A$ atoms,the Assertion is true.
Reason states that each represents $1 \ g$ molecule of oxygen,which is false because $16 \ g$ of $O_2$ is $0.5 \ mol$ and $16 \ g$ of $O_3$ is $0.33 \ mol$.

(C) To find the number of molecules,we calculate the number of moles for each option:
$(A)$ $11.2 \ L$ of $O_2$ at $\text{NTP}$ is $11.2 / 22.4 = 0.5 \ mol = 0.5 \ N_A$ molecules.
$(B)$ $8 \ g$ of $O_2$ is $8 / 32 = 0.25 \ mol = 0.25 \ N_A$ molecules.
$(C)$ $0.1 \ mol$ of $O_2 = 0.1 \ N_A$ molecules.
$(D)$ $2.24 \times 10^4 \ mL = 22.4 \ L$ of $O_2$ at $\text{STP}$ is $22.4 / 22.4 = 1 \ mol = 1 \ N_A$ molecules.
Comparing the values,$0.1 \ N_A$ is the smallest number of molecules. Thus,option $C$ is correct.

(C) The chemical formula of glucose is $C_6H_{12}O_6$.
One mole of glucose contains $6 \ mol$ of carbon atoms.
Therefore,$0.35 \ mol$ of glucose contains $0.35 \times 6 = 2.1 \ mol$ of carbon atoms.
The number of carbon atoms is calculated as $2.1 \times N_A$,where $N_A = 6.022 \times 10^{23} \ mol^{-1}$.
Number of carbon atoms $= 2.1 \times 6.022 \times 10^{23} = 12.6462 \times 10^{23} = 1.26462 \times 10^{24}$.
Thus,the correct option is $C$.

(D) The molecular formula of uracil is $C_4H_4N_2O_2$.
From the formula,one molecule of uracil contains $2$ nitrogen $(N)$ atoms and $2$ oxygen $(O)$ atoms.
Therefore,one mole of uracil contains $2$ moles of $N$ atoms and $2$ moles of $O$ atoms.
Thus,the correct option is $D$.

(A) The molecular formula of thymine is $C_5H_6N_2O_2$.
In one molecule of thymine,there are $2$ oxygen atoms and $2$ nitrogen atoms.
Therefore,in one mole of thymine,there are $2$ moles of oxygen atoms and $2$ moles of nitrogen atoms.

(C) The molecular formula of citric acid is $C_6H_8O_7$.
From the structure of citric acid,we can count the number of carbon atoms per molecule.
There are $3$ carboxylic acid groups $(-COOH)$,each containing $1$ carbon atom,and a central carbon chain containing $3$ carbon atoms.
Total number of carbon atoms per molecule $= 3 + 3 = 6$.
Therefore,in $1$ mole of citric acid,there are $6$ moles of carbon atoms.
In $n$ moles of citric acid,the total number of moles of carbon atoms is $6n$.

(C) The given structure is $2,3$-dimethylbutane.
Its molecular formula is $C_6H_{14}$.
In one molecule,there are $6$ carbon atoms and $14$ hydrogen atoms.
Therefore,in $n$ moles of the molecule,the number of moles of $C$ atoms is $6 n$ and the number of moles of $H$ atoms is $14 n$.

(A) The given structure is styrene $(C_6H_5-CH=CH_2)$.
Counting the atoms in one molecule:
Carbon $(C)$ atoms: $6$ (in benzene ring) $+ 2$ (in vinyl group) $= 8$ atoms.
Hydrogen $(H)$ atoms: $5$ (in benzene ring) $+ 3$ (in vinyl group) $= 8$ atoms.
Therefore,the molecular formula is $C_8H_8$.
For $n$ moles of the molecule,the number of moles of $C$ atoms $= 8 \times n = 8n$.
The number of moles of $H$ atoms $= 8 \times n = 8n$.
Thus,the correct option is $A$.

(C) The given structure is $but-2-ene$ $(CH_3-CH=CH-CH_3)$.
Counting the number of hydrogen atoms in one molecule of $but-2-ene$:
$CH_3$ $(3)$ + $CH$ $(1)$ + $CH$ $(1)$ + $CH_3$ $(3)$ = $8$ hydrogen atoms.
Therefore,$1$ mole of the compound contains $8$ moles of $H$ atoms.
Thus,$n$ moles of the organic compound will contain $8 \ n$ moles of $H$ atoms.

(A) The molecular formula of tear gas $(CCl_3NO_2)$ contains $3$ atoms of $Cl$ and $1$ atom of $N$ per molecule.
Therefore,in $n$ moles of tear gas,the number of moles of $Cl$ atoms is $3n$ and the number of moles of $N$ atoms is $n$.

(A) The chemical formula of mustard gas is $(ClCH_2CH_2)_2S$.
Each molecule of mustard gas contains $1$ sulfur atom.
Therefore,$n$ moles of mustard gas molecules contain $n \times 1 = n$ moles of sulfur atoms.

(B) At $STP$,$22.4 \ dm^3$ of any gas contains $1 \ mol$ of molecules.
$1 \ mol$ of $NH_3$ contains $6.022 \times 10^{23}$ molecules.
Each $NH_3$ molecule contains $4$ atoms $(1 \ N + 3 \ H)$.
Therefore,$1 \ mol$ of $NH_3$ contains $4 \times 6.022 \times 10^{23} = 2.4088 \times 10^{24}$ atoms.
Given volume $= 2.24 \ dm^3$,which is $0.1 \ mol$ of $NH_3$.
Number of atoms $= 0.1 \times 2.4088 \times 10^{24} = 2.4088 \times 10^{23}$ atoms.

(C) The molar mass of water $(H_2O)$ is calculated as:
$M = (2 \times 1.008 \ g/mol) + (1 \times 16.00 \ g/mol) = 18.016 \ g/mol \approx 18 \ g/mol$.
To find the mass,we use the formula:
$\text{Mass} = \text{moles} \times \text{molar mass}$.
$\text{Mass} = 0.25 \ mol \times 18 \ g/mol = 4.5 \ g$.
Therefore,the correct option is $C$.

(D) At $STP$ (Standard Temperature and Pressure),the molar volume of an ideal gas is $22.4 \ L \ mol^{-1}$.
Number of moles $(n)$ is calculated using the formula: $n = \frac{\text{Given Volume}}{\text{Molar Volume at STP}}$.
Substituting the given values: $n = \frac{0.448 \ L}{22.4 \ L \ mol^{-1}}$.
$n = 0.02 \ mol$.

(B) $1$. Calculate the molar mass of $CaCl_2$: $M = 40 + 2 \times 35.5 = 111 \ g/mol$.
$2$. Calculate the number of moles of $CaCl_2$: $n = \frac{222 \ g}{111 \ g/mol} = 2 \ mol$.
$3$. Since $1 \ mol$ of $CaCl_2$ contains $1 \ mol$ of $Ca^{2+}$ ions,$2 \ mol$ of $CaCl_2$ contains $2 \ mol$ of $Ca^{2+}$ ions.
$4$. The number of $Ca^{2+}$ ions is $2 \times N_A$.

(C) At $STP$,$1 \ mole$ of an ideal gas occupies $22.4 \ dm^3$ volume.
Given volume of gas = $4.48 \ dm^3$.
Number of moles $(n)$ = $\frac{\text{Given Volume}}{\text{Molar Volume at STP}} = \frac{4.48 \ dm^3}{22.4 \ dm^3/mol} = 0.2 \ mol$.
We know that $n = \frac{\text{Mass}}{\text{Molar Mass (M)}}$.
$0.2 \ mol = \frac{5.6 \ g}{M}$.
$M = \frac{5.6 \ g}{0.2 \ mol} = 28 \ g/mol$.
The molar mass of $N_2$ is $2 \times 14 = 28 \ g/mol$.
Therefore,the gas is $N_2$.

(D) At $STP$,the molar volume of an ideal gas is $22400 \ mL \ mol^{-1}$.
Number of moles of water vapour in $1 \ mL$ is $n = \frac{1 \ mL}{22400 \ mL \ mol^{-1}} = 4.464 \times 10^{-5} \ mol$.
The number of molecules is given by $N = n \times N_A$,where $N_A = 6.022 \times 10^{23} \ mol^{-1}$.
$N = 4.464 \times 10^{-5} \times 6.022 \times 10^{23} \approx 2.69 \times 10^{19}$ molecules.
Thus,the correct option is $D$.

(C) Let the mass of $O_2$ be $1x$ and the mass of $CH_4$ be $4x$.
Number of moles of $O_2$ $(n_{O_2})$ = $\frac{\text{mass}}{\text{molar mass}} = \frac{1x}{32}$.
Number of moles of $CH_4$ $(n_{CH_4})$ = $\frac{\text{mass}}{\text{molar mass}} = \frac{4x}{16} = \frac{x}{4}$.
The ratio of molecules is equal to the ratio of moles.
Ratio = $n_{O_2} : n_{CH_4} = \frac{x}{32} : \frac{x}{4} = \frac{1}{32} : \frac{1}{4} = 1 : 8$.